Solutions and Colligative Properties - Result Question 7-1
7. Freezing point of a $4 %$ aqueous solution of $X$ is equal to freezing point of $12 %$ aqueous solution of $Y$. If molecular weight of $X$ is $A$, then molecular weight of $Y$ is
(2019 Main, 12 Jan I)
(a) $4$ A
(b) $2$ A
(c) $3$ A
(d) A
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Answer:
Correct Answer: 7. (c)
Solution:
Given, Freezing point of $4 %$ aqueous solution of $X$.
$=$ Freezing point of $12 %$ aqueous solution of $Y$ or
$\left(\Delta T_f\right)_X =\left(\Delta T_f\right)_Y $ $\quad \left[\because \Delta T_f=T_f^{\circ}-T_f\right] $
$K_f \times m_X =K_f m_Y$
where, $m_X$ and $m_Y$ are molality of $X$ and $Y$, respectively.
or $ m_X=m_Y $
$ \text { Now, } \quad \text { molality }=\dfrac{\text { Number of moles of solute }(n)}{\text { Mass of solvent (in kg) }} $
$ n=\dfrac{\text { Weight }}{\text { Molecular mass }} $
$ \dfrac{w_X}{M_X \times\left(w_{\text {solvent }}\right)1}=\dfrac{w_Y}{M_Y \times\left(w{\text {solvent }}\right)_2} $
Given, $ w_X=4 \text { and } w_{(\text {solvent})_1}=96 $
$ w_Y=12 \text { and } w_{\text {(solvent})_2}=88 $
$ M_X=A $
$ \therefore \quad \dfrac{4 \times 1000}{M_X \times 96}=\dfrac{12 \times 1000}{M_Y \times 88} $
Thus, $ M_Y=\dfrac{12 \times 1000 \times M_X \times 96}{4 \times 1000 \times 88} $
$ =\dfrac{96 \times 12}{4 \times 88} \times \mathrm{A}=3.27 \mathrm{~A} \approx 3 \mathrm{~A} $
BETA
